Do we always have $\chi_{K/Q,x}=\pi_x$ if $K=\mathbb Q(x)$?

Question

If $K=\mathbb Q(x)$ for $x\in\mathbb C$. Do we always have $\chi_{K/Q,x}=\pi_x$ the minimal polynomial of $x$?

I am using the following definition: $\chi_{K/Q,x}$ is the characteristic polynomial of the application $a\mapsto ax$.

Answer 1

Yes.

Let $\pi_{x}=T^{n}+a_{n-1}T^{n-1}+\cdots+a_{0}$ be the minimal polynomial of $x\in \mathbb{C}$. A basis of $\mathbb{Q}(x)$ over $\mathbb{Q}$ can be chosen as $1,x,x^{2},\cdots,x^{n-1}$. Now we consider the matrix form of the map $a\mapsto ax$: \begin{align*} x\cdot (1,x,\cdots,x^{n-1})=(1,x,\cdots,x^{n-1}) \left( \begin{matrix} & & & -a_{0}\\ 1& & & \vdots\\ &\ddots& &\vdots\\ & &1&-a_{n-1} \end{matrix} \right), \end{align*} and this is the companion matrix of $\pi_{x}$, thus the character polynomial of the map is $\pi_{x}$.